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Physics 2053C – Fall 2001

Physics 2053C – Fall 2001. Chapter 4 Forces and Newton’s Laws of Motion. Forces. Newton’s Laws: If the force on an object is zero, then it’s velocity is constant. The acceleration = net force / mass.

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Physics 2053C – Fall 2001

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  1. Physics 2053C – Fall 2001 Chapter 4 Forces and Newton’s Laws of Motion Dr. Larry Dennis, FSU Department of Physics

  2. Forces • Newton’s Laws: • If the force on an object is zero, then it’s velocity is constant. • The acceleration = net force / mass. • Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first object. F = m a

  3. Fnet F1 F2 Vector Nature of Forces • Forces have: • Magnitude • Direction F = m a Net force and acceleration are parallel (m > 0). Vector Scalar Vector

  4. Common Types of Forces • Gravity • Forces between collections of atoms or molecules • Friction – opposes motion • Normal Force – opposes penetration • Tension – opposes separation or compression • Action-Reaction Pairs

  5. Examples of Forces: Tension M  T T Vertical Forces Tsin  + Tsin- Mg = 0 So: T = Mg/(2sin  ) W = Mg Horizontal Forces Tcos  - Tcos= 0

  6. M Examples of Forces: Friction N f = N  Forces Perpendicular to the Plane N - mg cos  = 0  N = mg cos   W = Mg Forces Along the Plane Mg sin  - N = 0 Mg sin  = N = Mg cos    = sin  /cos  = tan 

  7. M Examples of Forces: Friction N f = N  Mass remains on the incline up to a certain ; then it slides (accelerates) down the plane.  W = Mg s > k Static vs Kinetic Friction Forces Along the Plane  = sin  /cos  = tan 

  8. M Examples of Forces: Friction N f = N  Forces along plane Mg sin  - N = Ma Forces perpendicular to plane N – Mg cos  = 0  W = Mg

  9. M Examples of Forces: Friction N f = N  N - Mg cos  = 0  N = Mg cos  Mg sin  - N = Ma Mg sin  - Mg cos  = Ma Mg (sin  - cos  ) = Ma a = g (sin  - cos  )  W = Mg

  10. M Examples of Forces: Friction N f N  a = g (sin  - cos  ) If a = 0 then we can find . 0 = g (sin  - cos  ) cos  =sin   =sin /cos  = tan  W = Mg

  11. Top Block Bottom Block Nt Nb T f = Nt  mtg f = Nt Nt mbg T = 36 N  = 0.4 Examples of Forces & Acceleration 1 kg  = 20o 3 kg a

  12. Top Block Nt f Nt F = ma mtg T = 36 N  = 0.4 Examples of Forces & Acceleration 1 kg  = 20o 3 kg a Vertical Forces Nt – mt g = may = 0  Nt = mt g Horizontal Forces f = mt ax

  13. F = ma T = 36 N  = 0.4 Examples of Forces & Acceleration 1 kg  = 20o mb = 3 kg a Bottom Block Nb T f = Nt Vertical Forces Nb + T sin – mbg - Nt = mbay = 0  Nb = mbg + Nt - T sin   Nt mbg Horizontal Forces T cos  - f = mb ax

  14. T = 36 N  = 0.4 Examples of Forces & Acceleration 1 kg  = 20o mb = 3 kg a Horizontal Forces – bottom block T cos  - f = mb ax T cos  - mt ax = mb ax T cos  = mt ax + mb ax T cos  = (mt+ mb) ax T cos /(mt+ mb) = ax 36 cos 20/(1 + 3) = ax 8.45 m/s2 = ax Horizontal Forces – top block f = mt ax

  15. Next Time • Start Chapter 5. • Start getting ready for the quiz. • Please see me with any questions or comments. See you Monday.

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